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Can a plane land safely if it loses power during approach?
British Airways Flight 38 experienced a loss of power during approach to Heathrow and ended up crashing 270m short of the runway.
The aircraft was on glideslope when it lost power. The name "glideslope" suggests it should be able to glide to a safe landing. So why didn't it?

Suppose a commercial aircraft (e.g. 7X7, A3X0) lost all engines (say, due to birdstrike) shortly before landing. How likely would a safe landing on the planned runway be?

It would make sense for the approach angle to be such that a safe landing is very likely. So if not, why not?
EDIT: To clarify (due to @rbp's comment): I want to know how far is the typical approach angle from being able to glide in? And why is the approach angle not chosen so that gliding is possible?

Related: Why is "dragging it in" considered bad in small aircraft, but fine in larger aircraft?

"glideslope" doesn't mean the aircraft can glide to a safe landing. check your facts

@rbp Yes. BA38 clearly demonstrates that. I'm well aware that "glideslope" refers to the ILS beam, not "this is the slope the plane can glide at." I'm merely pointing out that the name is suggestive.

A standard “glide-slope” is 3–3.3°. 3° is 5.2% or 1:19, 3.3° is 5.8% or ~1:17.
A modern airliner will have lift to drag ratio between 17:1 and 20:1 in clean configuration and at best glide speed, which is usually somewhere in range 220–250 knots indicated, possibly even a bit more depending on weight. So it would be just barely able to follow the glide-slope if it was configured for best glide.
But on final approach it will at least need gear down and that increases the drag considerably. It will also already be slower than the best glide speed and have some flaps deployed and retracting them and accelerating to the best glide speed will take some altitude. Therefore if all engines quit on final, the aircraft is definitely landing short of the intended point.
For small aircraft I've read advice that if the runway is long enough, you should normally plan to use the middle part of the runway exactly so that you have a margin when engine quits on short final, accompanied with account of incident where it proved useful right there and another in the next chapter. However airliners need more runway, so they can't afford leaving much reserve on the near end, plus they usually follow the ILS or at least PAPI and those lead to fixed spot 1000 ft past threshold.

This varies for every plane and in every situation. Generally speaking every plane has a "Best Glide" speed. This is the speed through the air at which you will gain the largest ground cover for a given vertical distance traveled. Most planes have a published glide chart like this one for a PA-28. The published glide ratio is usually achieved at the best glide speed or or specified rates may be achived at a specific air speed.

(source)
You can also see that this chart is specifically for 0 flaps, gear up, and 0 wind. All of those factors will decrease the distance you can travel while gliding as they increase drag. The very need of flaps is to increase the approach angle without increasing the speed.
The glide slope is generally around 3 Degrees but more precisely it is at an angle which guarantees clearance from all obstructions that may be in the approach path so it may actually vary.
If you are on slope and you plane has a glide ratio that allows for an at or better angle you may be able to make the runway. However the term does not imply that it allows you to "glide" to a landing should the plane lose power.
On top of all that you will generally be landing into the wind, with that in mind if you have a strong head wind down the runway and you lose power you may may fall short due to an altered ground speed.
There is a nice debate on it here but smaller planes like a Diamond DA-20 has a glide ratio of 11:1 and a Cessna 172 has a glide ratio of 9:1 if memory serves. Various places around the internet seem to say bigger jets have in the 17:1 - 20:1 ratio range so they may even glide better than smaller planes. One difference with bigger jets is that they have larger speed ranges and if best glide speed is at 300KT the plane may be well below that on short final and by the time it pitches down to gain the speed it will be on the ground. Compare this to a Piper Warrior that has a 73KT best glide speed and a 70KT approach speed (sometimes I even come in at 75KT on a long runway in a warrior). It can be far easier to come to best glide in a smaller plane than a big jet. Keep in mind that on final approach you will have flaps down, gear down, and possibly slats down if your plane has them. That will greatly change what this chart would look like.  I These charts are mainly published so you know how far you can go if you lose an engine in flight.

Don't forget that flaps will change the glide ratio too.  The numbers that you are quoting are probably clean.

@mins, 3° is 5% or 1:20, so you need at least 20:1. The newest airliners may have that in clean configuration (no flaps, no gear) and fairly fast (well over 200 knots).

@JanHudec: Sorry I removed my comment before you clicked add. Thanks that makes sense.

@mins, I also expanded it to an answer now.

I will also amend my answer to include configuration variations.

Not all of those factors will necessarily decrease your gliding distance - a tailwind will increase it (as long as the tailwind is preexisting, as opposed to arising during the glide [which is generally bad]).

On "glideslope" doesn't mean the aircraft can glide to a safe landing.
The glideslope is a 3-degree approach path, which is considered the "normal" landing path.
Gliders, which don't have engines, and many warbirds are flown such that they can glide to the runway from any point on the approach.

Many aircraft are able to do an unpowered glide at the 3$^\circ$ angle that the typical (ILS) glideslope is. However that requires the aircraft to be in clean configuration (no flaps, no gear) and at a speed well above landing speed.
When the aircraft is configured for landing (flaps + gear down) it requires a fair amount of power to maintain on the glideslope.

It would make sense for the approach angle to be such that a safe
  landing is very likely. So if not, why not?

If the glideslope was steep enough to support an unpowered approach, then there would no margin to slow down the aircraft if it were approaching fast (one can't apply less power). It becomes difficult to stabilize the aircraft before landing. Also turbine aircraft need to be above idle power during the approach to ensure a fast response in case of a missed approach.
In case of a steep approach at near idle power, additional drag would be required to slow down. Most aircraft can only provide this by extending the spoilers. Unfortunately this also reduces the lift produced by the wing, something which is usually avoided on approach. A few transport aircraft such as the BAe 146/ RJ70 / RJ85 / RJ100 and Fokker F70/F100 are equipped with air brakes. They aid in slowing down the aircraft on steep approaches without compromising the lift. Steep approach modifications to the Airbus A318 includes changes to the spoiler logic such that they will extend only to 30% to prevent loss of lift. Propeller aircraft usually have less difficulty with steep approaches since the propeller can be used as an effective air brake. For these reasons the diversity of aircraft at London City airport (glideslope 5.5$^\circ$) is limited to a few models.
Currently there is an experiment running on Heathrow airport that features a increased glideslope (3.2$^\circ$ instead of 3$^\circ$) to reduce noise. Due to the steeper approach the aircraft require less power to maintain speed which relieves people living nearby of some noise.
This small increase in glideslope angle does not require special training for the flight crew or changes to aircraft but it may have a negative effect on the airport's capacity. Since the steeper slope reduces the deceleration capability of aircraft it might lead to an increase in go-arounds (e.g. because the aircraft's speed isn't stabilized at 1000ft, or the separation between successive aircraft is compromised).

I think a main difference between piston engines and jet engines is the time it takes for the jet engine to spool up.  Where a piston engine has near instantaneous power; a jet engine will take several seconds to produce full power.
It is this reason why a jet pilot prefers to have power in the mid range during an approach somthat if they have to perform a go-around the time it takes to spool up and produce full power is reduced.  If the angle was such that the engines were at glide power during the approach a go-around would be a much more dangerous.
If the glideslope were increased so its engine can be at idle during the dedcent, any failure that would cause the jet to land with less than full drag will inherently cause the jet to increase speed during the descent.  The pilots would be forced to use a shallower descent angle to keep the speed under control.  If the approach wasn't designed for a shallower approach it could cause the jet to get dangerously to terrain.

How does that answer the question about landing with all engines out during the final?

@mins It answers this question: It would make sense for the approach angle to be such that a safe landing is very likely. So if not, why not?

This question and its title has several non consistent standpoints, you answer this one, correct. Apologies. Actually this is completely true in the CDA.

The approach path of an airplane making powered vs unpowered (gliding) approach would be very different - normal powered approach is around 3 degrees steep, unpowered approach will be much steeper.  While a small piston engine plane can perform power-off approach and still be agile enough to power up on short final for a go-around, a big commercial jet can't do it.  Furthermore, the chance of a complete engine failure in a multiengine airplane is fairly small.  So it is much safer for multiengine aircraft to execute fairly shallow powered approaches (3 deg).

fairly small - it's vanishingly small!  Out of millions of flights, it's happened once?  I can't think of another example than BA038.

You're obviously right.  Whereas goarounds/aborted landings are not that rare.  Another point, I'm not a jet pilot, but I don't think you can really 'idle' a jet engine the same way as a small piston.  In any case, this is not something you would want to do.  And pax would not appreciate a diving approach either.

@Simon Well there's that and the two that the pilots shut down the only working engine, TransAsia 235 and British Midland 92

@Simon, what about the Dana MD83 flight 9J-992 at Lagos on Jun 3rd 2012. It crashed, killing all aboard, when after engines failed fairly early on approach and the crew focused on trying to restart them and failed to look for emergency landing spot.

Well that's 4 out of several hundred milion.


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What is the difference between Cloud Ceiling and Cloud Base?
By definition, Cloud Ceiling is the height above the ground or water of the base of the lowest layer of cloud below 6000 meters (20,000 feet) covering more than half the sky.
So does, Cloud Ceiling become Cloud base if it is covering less than half of the sky.

They are not the same thing.  Cloud base is simply the bottom "surface" level of the a cloud regardless of the sky coverage.  Where cloud coverage constitutes a ceiling, then the cloud base would be the same as the ceiling, but regardless, the bottom margin of any cloud is its base.

Simply put, a cloud base is the bottom of a cloud layer of any size. A cloud ceiling is the height of the first cloud layer that constitutes at least a broken layer.

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How fast can a large passenger jet descend and slow from cruise to emergency land?
On 17th April 2018, Southwest 1380, a Boeing 737-700, was cruising about FL320 when it suffered a catastrophic failure to engine 1.  It then took 5 minutes to descend to 10,000 feet, and I noted from the ATC recordings that the pilot said it would require a long approach to land at PHL.
Is this because the airplane could not bleed off airspeed fast enough for a short/straight-in approach? Is this the fastest that a large commercial airplane can descend and slow for an emergency landing?
I assume that, to achieve maximum descent and deceleration the remaining engine would be pulled to idle, and that landing gear would be extended at maximum safe speed, which IIRC is 250 knots?
Would it be reasonable to fly cross-controlled with an aircraft in this condition?  With no engine thrust requiring rudder input, is that safe at all speeds?  How much can it increase drag?
What are the speed limits for extending flaps, and how much do those affect drag?
(I can appreciate that in the case of Southwest 1380 the pilot in command, not knowing the extent of damage, may have been hesitant to do anything that would strain the airframe or controls.  So I'm curious as to how long an "unstressed" descent would take, and then how much faster one could descend by using the most aggressive techniques that a pilot would reasonably use in an uncompromised passenger flight in the same model.)

Related, related

32,000 ft down to 10,000 ft is 22,000 feet. 22,000/5min = 4,400 ft/min, that's a pretty good ear popping descent rate.  I try and limit my descents to 500 ft/min when flying my 4 seat prop plane.

@RalphJ: This is a set of hypothetical questions inspired by a recent accident.  It does not require or request any speculation about unknowns in the referenced accident or information that may become known in the future.  If you think any question inspired by an accident is off-topic, you should argue that in meta.

@fooot: Answers there are informative of part of my question, but this isn't just about getting to safe pressure altitude as fast as possible, but also how quickly a plane can slow for an emergency landing (and what factors might extenuate those procedures if the emergency is a catastrophic engine failure).

Note that 5 minutes is plenty of time. Passenger oxygen masks are required to last for 10 minutes. That’s for emergency descent down to 14,000 feet which is a breathable altitude if uncomfortable for some. They usually do 10,000 feet because it is comfortable for most people.

A 5-minute descent from FL320 to 10,000 feet is pretty quick. As noted in the comments, the descent rate is 4,400 ft/min. For reference, during final approach descend rate is 300~500 ft/min in a small airplane, 500~700 ft/min on an airliner. A typical descent from cruise altitude is performed at 1200~2400 ft/min.
Getting down fast
A "dive bomb" flight-simulator-style descend is not practical if you intend to land. After all, you have to arrest the descend and bring the aircraft to a stable state. You may very well break the plane when trying to pull out of the dive.
Now, what is the fastest way to lose altitude? According to the Boeing 737 QRH, the Emergency Descent checklist (which is conveniently listed on the first few sections) says...

(...) Spreadbrake...FLIGHT DETENT
(!) If structural integrity is in doubt,
limit speed as much as possible and avoid high maneuvering loads.
Set target speed to Mmo/Vmo. (...)

In other words, deploy the speedbrakes, then pitch down to max airspeed.
Note that, however, this procedure is only used when it is absolutely necessary to get down as quickly as possible. The Boeing 737 FCOM also contains the following, in bold letters:

WARNING: Use of speedbrakes at speeds in excess of 320 KIAS could
result in severe vibration, which, in turn, could cause extreme damage
to the horizontal stabilizer.

Landing gear & Flaps
Deploying the landing gear is not good because:

The maximum flying speed with gear extended is 320 knots, while Vmo is 330 knots. Therefore, if you descend with the gear extended, you will lose altitude slower.
The maximum flying speed for extending the landing gear is 270 knots. Therefore, if you wish to extend the gear, you will have to slow down to 270 knots first, then lower the gear handle, wait for the gears to be locked, then pick up speed again.
Flying with the gear extended introduces vibration to the airframe. Deploying speedbrakes already introduces vibration. A damaged engine further adds vibration.
You need the gears to touchdown, so better protect them in the landing gear bay.

Extending the flaps is even worse because the maximum speed for flying with flaps is even lower. The maximum flap extension altitude is also 20,000 ft.
Cross control
Cross-control can probably lose altitude even quicker, but it is not a maneuver you would do on a passenger airliner. When crossing control, the gravity vector is not down, but sideways - meaning passengers will be forced to the side of the seat, a sensation they are not used to. Furthermore, the passengers sitting on the window side will likely perceive the attitude as extreme bank angle and think that the plane is going to crash! The last thing we want in this scenario is a cabin with panic passengers.

I gather from news reports that some large number of the passengers thought the plane was doomed and they were going to die.  The one who did die was partially sucked out the broken window.  (Now I'm wondering if, with a full awareness of that condition, slipping with the exposed side of the cabin forward could offset the decompression force through that hole....)  In any case, if I were PIC and I had to choose between saving the craft but making the passengers think they're going to die, or keeping them calm but actually killing them in a failed landing, I'd go with the former ;)

Speed and vertical speed are not proportional in this case, so the slower speed with gear in point 1 does not actually mean anything. The vertical speed is given by loss of energy to drag, so if the drag with gear is sufficiently higher, the descent rate could be higher with gear at 320 knots than without it at 330 knots. The second point is a good reason not to extend it anyway though (and it is not part of the emergency descent procedure).

This is a very good answer. I’m only a single engine pilot but I can’t imagine willingly going cross-controlled on a twin with one engine out. Especially when you have fuselage damage,  I would want to keep the forces as normal as possible.

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Are control inputs different during autorotation?
My understanding is that during power-off autorotation the main rotor is still coupled to the tail rotor, and the flight controls execute the same functions as during power-on flight.  Two questions:

Do any (heavy) helicopters have power-assisted controls?  If so are those difficult to handle in the event of total power failure?

How is the input during autorotation different from the input for the same attitude under power?  E.g., during a given glide and flare to land are the rotor blade angles of attack and swashplate orientation identical to when the rotors are powered?

Yes and yes.
A certification requirement of helicopters with powered controls is that they have a manual reversion so that control can be maintained in the event of a power failure.  However, the controls can be really heavy with no power and accidents have occurred when hydraulics have failed at critical moments or pilots have lifted off with hydraulics off and have been caught by suprise with the effort needed to maintain control.
The control inputs during autorotation are the same.  For any given glide, or flare, the lift required is the same powered or unpowered.  Lift is proportional to the speed of the blades, which is independent of powered or unpowered, and the angle of attack.
During autorotation, the energy (power) going into the rotors is derived from the airflow coming up through the disc. In this sense, an autorotation is still powered but the power comes from the airflow rather than the engine.
During autorotation, since the airflow is from below, rather than above, the pitch required is lower for the same angle of attack to generate the same amount of lift.  So, the swashplate is lower in autorotation but lowering collective still decreases lift and raising it increases lift.  The primary difference is that rather than controlling lift and power, the collective controls rotor speed since increasing pitch increases the angle of attack which increases lift and therefore increases drag, slowing down the rotor.  Lowering the collective has the opposite affect.
The cyclic control is the same in all respects (you tilt the rotor thrust in the direction you want to go) since it increases, or decreases the pitch, and therefore the angle of attack, cyclically.  The movement of the swashplate about the X and Y axis is independent of the position of the swashplate in the Z axis.
The pedals work in the same way too.
In summary, the only difference in the controls is the vertical position of the swashplate.

Do you want to have translational motion during autorotation?

@TomMcW Absolutely, although it's possible to autorotate most of the way with zero airspeed.  You need it at the bottom though to be able to flare, to increase rotor RPM so that you can use the collective to generate lift and cushion the landing.  You change from converting height into rotor RPM to converting speed to RRPM.  Most autorotations are flown at max range speed, or minimum descent speed or some other nominal figure, typically at around 60 kts since this is where most helicopters require the least power.

'"A certification requirement of helicopters with powered controls is that they have a manual reversion so that control can be maintained in the event of a power failure."' This is incorrect. There are many helicopters with hydraulic only, fully irreversible flight controls: if all hydraulic power fails, the helicopter cannot be controlled.

Image source
In heavy helicopters such as the CH-53E Super Stallion, the swash plate is actuated hydraulically only, there is no manual reversion and if all hydraulic power is off, the helicopter cannot be controlled. That is why there are redundant hydraulic systems, powered by pumps mechanically connected to the rotor transmission: as long as the rotor turns, the pumps provide hydraulic power. Loss of all hydraulic power is considered highly unlikely.
The main hydraulic actuation cylinders are located at the swash plate and at the tail rotor, and the cable run to the flight controls is of considerable length, creating some friction and lack of precision. Also, there is a mechanical mixer installed which provides coupling between collective-pedals-cyclic: when collective is increased, the mixer provides extra tail rotor input without the pilot having to move the pedals. Without any additional measures, the mechanical mixer also creates cross coupling between flight controls which would further complicate precision control: it is a big ship!
In order to make the stick forces simple and light, there is a second hydraulic actuator at each flight control, just behind the cockpit: they are irreversible, and the AFCS system can independently move these and add autopilot inputs. The feel characteristics are as follows:

With AFCS ON, only artificial feel forces (including trim) are felt plus the friction of the short run to the AFCS system. If the pilot moves a control, the mixer does its work without any other flight control being back driven.
With AFCS OFF, the higher friction of the run to the swash plate primary actuators is felt, plus back driving forces from the mechanical mixer.

During autorotation, the coupling between the swash plate and the flight controls is exactly the same as when in powered flight. However, there is considerable difference in the feel and in aircraft response with AFCS ON or OFF. Both conditions are modelled in flight simulators and the pilots are experienced in controlling the ship in either condition.

Heavy helicopters tend to be hydraulicly controlled to such an extent that there no longer is a physical link between the controls and the blades (like big jets). You're just opening and closing hydraulic valves one way or another. So... no hydraulics, no controls. The hydraulic pumps are generally driven by the main gearbox so that you will always have hydraulic pressure (main and backup) as long as the rotors are turning faster than a minimum speed. Usually you will fall out of the sky before that rotor speed falls below that threshold.
There is a rotor speed that if it falls below that rpm, you will have lost the ability to get it back up and will have lost the ability to auto rotate.

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Why don't helicopters always takeoff from a hover out of ground-effect?
I've read enough crash reports to see that helicopters discovering too late in takeoff that they lack the power to fly out of ground-effect is a serious problem.
I am wondering why the following isn't standard takeoff practice: Establish positive climb rate out of ground effect before beginning any translational flight.  Failure to climb out of ground effect would be cause for immediate ATO, and there would be this nice landing zone to abort into!
(Only in exceptional emergencies, like when a mountain takeoff gives the rotor-craft a clear glide-path into denser air, would it be reasonable to ignore this procedure.)
(I note that I am assuming that translational flight out of ground effect is always more demanding than hovering.  I vaguely remember that this is not the case, but I don't remember why, and it seems counterintuitive since translational motion always establishes a retreating blade with lower lift that then needs a higher angle-of-attack and more power to maintain lift.  So if the helicopter can't climb under full power with no translation, it would not be able to once it tries moving forward.)

I don't think I can attempt a full answer on this, but I believe that the reason they translate to forward flight has to do with the potential of overheating the engine during hovering. Forward motion gets air flowing over the engine, helping it to control temperature.

I'm a bit late to the party, but there is some related information here.

Forward flight is much more efficient than hovering.  As airspeed builds, lift increases from "translational" lift as the air moves more horizontally over the disc.  Since the relative airflow is more horizontal, the angle of attack for a given pitch angle is increased.
The vortices and turbulence move behind and down from the helicopter so undisturbed air is drawn over the disc to be accelerated downwards to produce lift.  Lift is more efficiently produced by accelerating a large mass by a small amount rather than a small mass by a large amount.
It is more efficient to remain in ground effect and slowly increase speed to gain translational lift than to increase power to HOGE (Hover Out of Ground Effect) and attempt to translate.
The problems you refer to are where an unwary pilot attempts to transition from HOGE whilst allowing the aircraft to climb or attempting to maintain height.  Without enough power, the loss of ground effect before the benefits of translational lift are gained will result in a rapid descent back to earth (as vortex ring state or settling with power is pretty much impossible to avoid) or you overpitch and drag the RRPM (rotor RPM) down so low that you cannot produce lift or even worse, stall the blades.
Bottom line, it takes more power to establish positive climb out of ground effect than to gain translational lift in ground effect.

Thanks, I had forgotten about the benefits of translating the disc into "smooth" air.  However, at some larger translational airspeed the translational lift does begin to fall as increasing (retreating) areas of the disc are unable to maintain positive relative airspeed, right?  Or is that always outside of the flight envelope of a helicopter?

The benefits of translational lift do drop off but not in a linear fashion. Remember that as the blades flap up (advancing side) and down (retreating side) the angle of attack also changes. The blade(s) flapping up experience a reducing angle of attack because the relative airflow is more from above and the blade(s) flapping down experience the opposite. Therefore, there is a natural tendency for the advancing blade to generate less lift and the retreating blade to generate more. Dysermmetry of lift becomes a problem as VNE is exceeded.

@Simon two things you might want to addd: 1. just pulling into a hover requires a fair amount of power, and will help the pilot determine whether she has enough power to take off (2) there's always the running take-off

You state that: 'As airspeed builds, lift increases from "translational" lift as the air moves more horizontally over the disc. Since the relative airflow is more horizontal, the angle of attack for a given pitch angle is increased.'    Well, the opposite is true: for a given pitch, the more parallel the relative airflow is to the rotor disk, the less the AoA...

A helicopter takeoff profile is governed by having enough margin to autorotate to a safe landing in event of engine failure, not by more efficient flight due to ground effect. Margin consists of both forward speed and altitude.
Helicopter manufacturers publish altitude vs airspeed charts which prohibit operation above a survivable altitude until sufficient airspeed is achieved. Height without speed, as you suggest for use in a HOGE profile, results in high speed vertical impact on engine failure. Staying in ground proximity until there is sufficient forward velocity to initiate autorotation means vertical impact speed stays low with loss of power.
It just happens to be that a survivable altitude is also in ground effect, so efficiency is a side benefit of a typical takeoff profile rather than a driver.


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How close to the envelope and the ideal was US1549 on ditching in the Hudson?
I assume that for "water landing" an airplane should be configured for the lowest possible stall speed, and then flown to that speed before impact to minimize the energy to dissipate on impact.

What is that configuration for a plane like the A320?  E.g., full flaps?  Was UA1549 so configured?
How close did UA1549 get to the stall speed before impact?  If I understand this answer then the A320's fly-by-wire protection would not allow a pilot to stall the airplane, meaning that the Pilot Flying could pull full aft to nail that speed (without either pilot calling airspeed to avoid a stall, which I assume would be the procedure for a plane without a self-protection computer).
Was the ideal pitch for a power-out water landing known by the plane and/or the pilots?  Normally we "pitch for speed," and it would seem an unlikely coincidence for the stall pitch to match the ideal impact pitch.  Which pitch was higher?
If impact pitch was higher did the Pilot Flying have to account for this, flying above stall speed in order to permit an increase to impact pitch immediately before impact?

Just a note, stall speed doesn't mean the plane stops flying. Often right at "stall speed" if you pull back hard to try to slow down you will balloon and impact the water harder (because your sink-rate at impact will be higher). See Why is ditching speed higher than stalling speed?

@RonBeyer: Makes sense, and based on the answers to that question I deleted the last point of this one because, if for no reason than the difficulty in judging altitude above water, it sounds practically impossible to execute a flare to drop speed below Vso and then lower to impact pitch to "grease onto" the water with a sink-rate lower than the steady minimum glide sink-rate.  Not to mention that minimum steady sink rate might be at an airspeed above Vso.

The configuration of the A320 for ditching with no engines running is CONFIG 3. I don't know how familiar are you with the flap configurations on A320, so I will start from the beginning. The flap lever has five positions: 0,1,2,3 and FULL (two configurations correspond to position 1-> CONFIG 1 and CONFIG 1+F).
In the CONFIG 3, which is the preferred method for ditching, the slats are extended to 22 degrees and flaps are down by 20 degrees. Config FULL, has slats down to 27 degrees and flaps down to 40 degrees. Look at the difference between flap deflection in CONFIG 3 and CONFIG FULL, that's 20-degree change in flaps. The 20-degree change will make a lot of impact on the drag of the aircraft and its vertical speed at the moment of impact. Therefore, CONFIG 3 is preferred. I must note that CONFIG FULL will not decrease the stall speed by a large amount: from my experience, it will decrease it by about 5 knots or a maximum of 10 knots. But definitely not more than that. The US airways flight 1549 had flaps down to CONFIG 2, below the ideal state.

If the ditching is performed with engine thrust, the suggested configuration for ditching is then, obviously, CONFIG FULL because we can trade engine power for the lowest possible approach speed which, for flight 1549, was not the case.

Flight 1549 was pretty close to the stall speed. On the video below you can see the approximate speed at the moment of impact. The exact touchdown speed was 125 knots.

With dual engine failure on Airbus A320, the aircraft will revert to alternate law, but due to the captain's fast reaction to turn on the APU, aircraft remained in normal law, assisting the pilot to not stall the aircraft. This does not mean that if the pilot pulls all the way back at such low altitude that no stall will occur. Upon landing/touchdown normal law goes gradually out and the FBW system introduces a nose-down pitching moment, so the pilot will not balloon the aircraft. In the last seconds before touchdown, the computer gradually returns to direct law, giving the pilot more feeling over the pitch. In this abnormal scenario, it is difficult to judge exactly what the FBW would do since this "reverting the pitch control to direct law" thing is characterized by the FLARE mode on the FMA, and I am not sure if the FBW system met requirements for that mode. If normal "reverting to direct law" occurred then the pilot would have been able to stall the aircraft just before the touchdown. If it did not occur, due to all abnormalities that occurred to the aircraft, the pilot would not be able to stall the aircraft.
The plane definitely does not know the correct pitch for ditching, neither what the crew was attempting to do. In ditching, the best pitch attitude for A320 is 11 degrees. In the answer that you linked, I can see that for 11-degree pitch attitude, the aircraft at the given weight is descending at -1 degrees, which makes the angle of attack equal to 12. Look the accident flight, angle of attack was 13 degrees, by the same analogy. That means that the aircraft was slower than ideal although the speed was 7 knots higher (due to the takeoff weight of the aircraft). The point is, given its weight the aircraft should have flown faster (than 125 knots) and the pilot should have pitched the nose of the aircraft higher, to match the preferred attitude of 11 degrees.
Stall pitch should be higher than the actual pitch for the ditching.

On point #2: On an abnormal approach like in this scenario does the FBW have any clue what's going on?  If it believes the radar altimeter then it knows it's approaching ground, but it sounds like, at least based on what you know, it's unclear what mode that would put the FBW in, and therefore how much authority the pilot would have to "push the envelope?"

FBW modes are much more complex since their operation depends upon dozen of systems and not just one.
Look at this question https://aviation.stackexchange.com/questions/52387/how-do-airbus-planes-flare-in-normal-operation?rq=1
Probably the pilot wouldn't be able to stall the aircraft.

Based on the UA1549 cockpit recording it sounds like it would have been far more helpful if the FMA had kicked in instead of the nagging GPWS!  (It did not.)

Note that the pilot was pulling back on the sidestick at touchdown trying to get the plane to a higher pitch angle (10-12° is Airbus recommended angle) but the system stopped it at a 9.5° AoA due to a phugoid prevention system.


text:
Does MMO (limiting Mach number) assume any maneuvering limits?
This answer to a question about emergency descents notes:

[Y]ou can accelerate descents by banking the aircraft
  into a turn. This allows you to increase angle of attack, thus
  increasing drag to keep speeds down, while rotating the lift vector
  away from the vertical so it's not working against you. The
  combination allows for higher descent rates with the same MMO/VMO
  limits.  Some operators train for this, but my impression is not many....

Banking seems like an obvious way to increase drag and accelerate descent when the only objective is to get to a lower altitude (e.g., due to rapid decompression).  So does forward slipping.  Is either ever indicated in manuals or emergency checklists?
I was wondering if such maneuvers might actually be contraindicated: Isn't MMO determined by the stress limits of the airframe?  If so, does MMO (in theory, if not in practice) assume some limit on angle of attack, and would MMO decrease at high angles of attack or slip?

The problem is that in a deliberately entered spiral dive it is very difficult to keep things in check.  In an idle descent, pitch is your only speed control (well... really, in the end, pitch is your speed control all the time; any glider pilot understands that), and in a spiral dive trying to slow down by increasing pitch is trickier because a lot of the pitch input just steepens the spiral and you just pull more G instead of slowing down.  Which is the whole reason spiral dives are to be avoided in the first place, especially in the IFR world.
In a jet in an emergency descent you will generally dump flight spoilers with the engines at idle and push over to accelerate to MMO/VMO.  An attempt to maintain MMO/VMO in a spiral dive turn would probably end poorly.  If you are diving straight at VMO, spoilers out, and want to get down faster, depending on the airplane you might get a faster descent by slowing to gear extension speed and extending the gear, but I don't think most manufacturers recommend that and it's not something I was trained to do (on RJs).
Sideslip doesn't come into it as sidesliping swept wing aircraft deliberately is very dangerous (it doesn't take much skid/slip to make a swept wing airplane want to roll on its back), and nasty dutch roll situations can result.  In any jet, once you are off the ground, the yaw damper system takes care of adverse yaw and any random disturbances, and you never touch the rudder pedals unless an engine quits.  It's generally feet on the floor after the initial departure profile until final approach.
In any case, in an emergency descent you are way above maneuvering speed so pulling too hard will fold the wings before stalling AOA is reached and trying to get more descent rate by spiraling would be suicidal.

Isn't MMO determined by the stress limits of the airframe?

No. It is usually determined by the onset of “Mach tuck”, the transsonic loss of lift caused by the flow speed over the wing becoming supersonic in too large region.
It may be also determined by the TAS limit due to onset of aeroelastic flutter, because Mach number follows true airspeed more closely than indicated airspeed.
Both effects are pretty dangerous. The pitch-down moment associated with Mach tuck might put the plane in irrecoverable dive, at least until it gets much lower to warmer (higher speed of sound) denser (higher drag) air, but in the meantime it will likely exceed some other limit and might get damaged.
And aeroelasic flutter will break the plane apart if left to continue for some time.

If so, does MMO (in theory, if not in practice) assume some limit on angle of attack, and would MMO decrease at high angles of attack or slip?

Actually I believe it would. If the wing is forced to generate more lift, the flow speed above it will be higher and it will separate earlier.


text:
Can weight reduce a plane's glide ratio?
The FAA says glide ratio doesn't change, regardless of weight, here on page 5-7.

Glide ratio is not affected by weight because,
while a heavier glider sinks faster, it does so at a greater
airspeed. The glider descends faster, but covers the same
horizontal distance (at a higher speed) as a lighter glider with
the same glide ratio and starting altitude.

However, a friend suggested to me that this is BS, and if you overload a craft above its operational limits, its glide ratio will plummet. In fact, he's confident that there's a sharp decline in glide-ratio from the point of going over the maximum weight limit. I don't know if this is true, but am interested to learn.
Question

The question is, quite simply, is the FAA inaccurate/wrong on this point? Can you reduce glide ratio by increasing the wingloading of a craft, as my friend suggested?
My Attempts at Research
I tried to look for polar curves that show the performance of overloaded craft, but couldn't find any' only curves within operational limits. I have no intention of testing something so dangerous myself, of course, but I do want to know the answer. I have heard anecdotes of people going well above the safety limits of their craft, for things such as world records.
I'm hoping wise people here can teach me a bit about the facts, and whether the FAA's statement is technically wrong as was suggested.
Additional Points:
He pointed out cargo planes have extremely severe limitations, where exceeding their weight limits will often be catastrophic.
There was a world record flight, I forget the plane and pilot, but I recall it was apparently loaded up to almost twice it maximum weight with fuel tanks.
Glider weight limits I saw seemed more related to landing and taking off than to concerns about performance in the air.
I've seen a couple of people state that glide ratio will only improve with a higher weight (very slightly), up till the point where A, the materials break from being overloaded, or B, the optimum speed will reach the Vne and become disabled for that fact.
Notably, even if glide ratio does remain consistent with more weight, landings, takeoffs, and maneuvering become more difficult. It's also harder to make use of thermals.

Tape a rock to a paper airplane and see how it glides.

@Jim By this you mean the FAA was wrong, and weight does have a huge impact on glide ratio? It is surprising they said it isn't affected by such, which is why I wanted to ask about this.

I don’t think the FAA is talking about “beyond operational limits”

@Jim It seems strange they make no mention of that. A couple of people on other sites have tackled the question, and believe that glide ratio will continue to remain static, and the craft will break from strain before that's likely to change. You could technically reach a point where optimum speed exceeds Vne, but they have stated that the increase in optimum speed is the square root of the change in weight, so that's very difficult to achieve. I'm hoping someone can go more deeply into this and confirm what the case is.

The glide ratio is a constant while operating within the normal operating conditions of the aircraft.  Once you go outside those limits, then you likely have either a structural failure or a stall. If the FAA were interested in what happens outside the limits, they would have stated that in the question.

@jwh20 The FAA manual is not answering a specific question, on that page. It is teaching about gliding principles, and makes no clarification I could find that it means what you suggest. Indeed, it has been negligent in this if that is the case, as several pilots from a couple of sites have claimed those principles are accurate, as I described, that this is the case beyond operating weight limit. I would appreciate it if someone knowledgeable about aerodynamics can explain in some detail about what the case is.

In light of your additional clarification here, you may want to review your original question and simplify it.  It seems you are specifically interested in the aerodynamics of lift vs. load.  It's hard to discern that from your question.

@jwh20 Thank you, but I'm not sure I'm skilled enough to work out how to simplify the question. The whole question arises from a specific and messy point. I will make an effort to simplify it.

The original question seems fine to me.  Please do not alter it in any way that would invalidate an existing answer.  Thanks!

@quietflyer I'm not criticizing the question, the OP was.

However, a friend suggested to me that this is BS, and if you overload
a craft above its operational limits, its glide ratio will plummet. In
fact, he's confident that there's a sharp decline in glide-ratio from
the point of going over the maximum weight limit. I don't know if this
is true, but am interested to learn.

Your friend is wrong.  For that to happen, there would have to be a sharp decrease in the lift coefficient, and/ or a sharp increase in the drag coefficient, associated with any given angle-of-attack, once the maximum weight limit is exceeded.  This doesn't happen.
Background info--
Glide ratio is the ratio of Lift to Drag, which is also the ratio of Lift coefficient to Drag coefficient.
In what follows we'll assume we can ignore the slight changes in Lift coefficient and Drag coefficient-- in particular the slight decrease in Drag coefficient-- associated with a slightly higher Reynold's number, as airspeed is increased at a given angle-of-attack.  We'll also assume we are staying firmly in the subsonic range, so we can ignore Mach-related effects.
For any particular aircraft in a particular configuration, Lift coefficient and Drag coefficient depend entirely upon angle-of-attack.  As long as angle-of-attack is constant, these coefficients are constant.
Therefore the best glide ratio relative to the airmass, which also the best glide ratio relative to the ground in still air, always occurs at the same angle-of-attack regardless of weight, for any particular given aircraft.  (Optimizing the glide ratio relative to ground in various headwind or tailwind conditions appears to be beyond the scope of this question, so we'll assume that we're always talking about the still-air glide ratio, or the glide ratio relative to the airmass.)
For any given angle-of-attack, the effect of an increase in weight can be  represented as follows-- to calculate the new horizontal speed and the new sink rate, take the polar curve (graph of horizontal speed versus sink rate), draw a line from the origin of the graph through the point representing flight at the best glide ratio at the lower weight, and then extend the line further in the same direction, so that the new length is increased over the original length, in proportion to the square root of the ratio of the new wing loading to the old wing loading.  Both the horizontal and the vertical speed (sink rate) will be increased in proportion to the square root of the ratio of the new wing loading to the old wing loading.  And so will the airspeed.  So the glide ratio will remain unchanged.  (Note that the airspeed is actually represented by the length of the diagonal line we've drawn, from the origin of the graph to the (horizontal speed, vertical speed) data point.  For all practical purpose at reasonable glide ratios (small glide angles), this is also the same as the horizontal speed, but this approximation breaks down at extremely poor glide ratios (large glide angles).)
(Disclaimer: we're assuming that we're not breaking, or hugely deforming, the aircraft with the heavy load.  It's not obvious that conventional aircraft would tend to deform in any way that significantly changes the glide ratio, before the point of structural failure.  In flex-wing hang gliders, the flexible aluminum tubing comprising the leading edges of the wing does deform under heavy load in a way that changes the shape of the rest of the wing and greatly increases twist and washout and decreases glide ratio, but that's obviously beyond the intended scope of the FAA's comment that is the subject of this question.)

Glider weight limits I saw seemed more related to landing and taking
off than to concerns about performance in the air.

Actually, glider weight limits are primarily dictated by concerns about the integrity of the structure under heavy load.  You can imagine what would happen to the main wing spar, or the wing-fuselage attachment points, if you overload the glider severely and then "pull" several G's in a steep turn, or hit severe turbulence that imposes several extra G's of gust loading.
Links to related ASE questions:
Why is the L/D ratio numerically equal to the glide ratio?
Can we show through simple geometry rather than formulae or graphs that the best glide ratio occurs at the maximum ratio of Lift to Drag?
How are the glide polar and L/D ratio charts related?
Why would a glider have water ballast? If it is trying to stay aloft without an engine, wouldn't it be better to be as light as possible?

Nice write up @quiet flyer . The rub is most GA aircraft publish only one best glide speed, incl in the emer checklist for engine out. So, technically, on a C-172S, using the V is proportional to sqrt(W) rule, we should fly as much as 5 knots off of 68 KIAS, for optimum L/D. Fortunately, L/D is only slightly decreased over that relatively small range and, as you point out - better to pay attention to other things such as wind direction, suitable landing sites, checklists, etc. BTW, the C-172 POH max glide chart has three thick lines plotted, presumably to depict some of this scatter.


text:
What is the effect of uneven ground on ground effect vehicles?
I understand ground effect as far as Wikipedia explains it. There are two ways to ask my question;

A ground effect vehicle flies along and encounters an obstacle (say a big wave) that's maybe about half the flying height. What will this feel to the craft? My assumption would be a short moment of more lift, which would feel bumpy at best and could damage the craft at worst.
How choppy may the ground be? From the articles about Ekranoplans and similar concepts I read all these vehicles seemed to be designed for inner seas like the Caspian Sea or the Wattenmeer, not for the Pacific. This may be a coincidence or simply wrong. But can we quantify somewhat how seaworthy such a craft would be, or what were the most extreme conditions in which a ground effect vehicle has operated?
When I try to answer the question myself, the one 'killer' condition I can come up with is when the ground rises faster than the craft can gain height (duh). Is this correct, are there others?

I've put this in a list so you can see where my misconceptions might be when addressing the underlying question: the ground in ground effect....

The ability of a ground effect (air)craft to able to fly with (small) perturbations in height from the ground is called height stability. In general, height stability is achieved if the derivative of the lift coefficient with increasing height is negative. In this case, a decrease in height (due to waves etc.) causes an increase in lift. As a result, the aircraft is returned to its original height due to the increased lift (with the opposite happening when height is increased).
As the ground effect increases as the height decreases, designing the ground effect aircraft with height stability is not very difficult and most of the ground effect aircraft designed and operated have height stability. A good example is the Lun Ekranoplan, which returned to its height after the height was altered by approximately 0.5m (due to missile launch).

"Lun Ekranoplan" by Soviet Navy - Stock Archives of Soviet Navy. Via Wikipedia.
Also, the Sea state limitations are a scale phenomenon — the larger the ground effect aircraft, rougher the seas it can handle. The Russians (Soviets) have the most experience in operating ground effect vehicles and according to them, sea state limit for a 500-tonne aircraft is around 2.5 m and the safe operating heights in terms of mean wave heights is given by,
$h  = \frac{1.54 H_{\frac{1}{3}}}{2} +  0.1  c$
where,
$h$ is the vertical height measured from the mean wave height,
$c$ is the chord,
$H_{\frac{1}{3}}$ is the average of the 1/3 highest wave.
When this height exceeds the ground effect height the craft operates as an aircraft at a lower efficiency.
The main problems in case of operating a ground effect aircraft in choppy waters (as you noted, they are mostly operated in 'calm' seas) are twofold:

Takeoff and landing are dangerous in rough seas, due the impact loads on the hull and wings. This is the most important limitation for operating in rough seas.
Stability — The wing is already unstable, and near the ground in high lift region, it is more so; this is the reason most of the ground effect vehicles have proportionally larger tail surfaces compared to the conventional aircraft. A good example is the Soviet -90 'Orlyonko', which had tail surface nearly 50% of that of the wing.

"A-90 Orlyonok" by Kaboldy - Own work. Licensed under CC BY-SA 3.0 via Commons.
The main problem with waves (or in general, variations in height) is that the pitch stability changes with height. This results in the requirement of a large amount of control power in order to maintain trim — another reason for large tail control surfaces. Generally, the aircraft is stable in roll (the dipping wing generates more lift and the aircraft corrects itself).
Another thing to note is that the maneuverability of the ground effect aircraft is pretty limited if it's unable to fly in OGE (Out of Ground Effect). Soviet GE aircraft could fly in OGE for clearing obstacles.
References: Wing in Ground Effect Craft Review by Michael Halloran and Sean O'Meara

for clarification: the height h is the lower safe height, the craft operates in a band between h and the ground effect height that's roughly the wingspan? (so when h > wingspan, it's an inefficient plane)? What does "the wing is already intsable" mean? What does OGE mean? This answer looks great, I just want to understand a few more details!

@mart When h>wingspan, the aircraft operates just like any other conventional aircraft. The advantages due to ground effect are lost. Compared to ground effect aircraft, they are inefficient. Wing alone configuration is unstable. That is the reason for having a horizontal stabilizer- to make the aircraft stable. OGE means Out of Ground Effect.


text:
Why is max endurance different for jet and props?
Max endurance speed is not the same for gliders, propeller planes (also for turbo prop, right?) and jet planes.
While for for jet planes max endurance is at the minimum drag speed ($V_{md}$), for propeller planes and gliders it is the speed where minimum power ($V_{mp}$) is required. While I found hundreds of posts explaining that the speeds are different because the formulas mentioned above are the way they are, I could not quite figure out why they are different. (I am not an engineer and could not quite follow some of the more lengthy, mathematical explanations.)
In this question the author states:

In case of propeller aircraft, the fuel flow rate is proportional to
  the power produced. Hence, the maximum endurance occurs at a point
  where the power is minimum. For (turbo)jets, the minimum fuel flow
  occurs when the thrust is minimum. Hence the maximum endurance occurs
  when the L/D is maximum.

Of course this explains why max endurance is different, but it creates the new question of why propeller engines behave differently from jet engines.
At the end of the day both engines have the same task: Convert chemical energy to kinetic energy.
If I imagine a jet plane and a propeller plane with parking brakes set and a given throttle input they will both produce a lot of thrust, but zero power. Advancing the throttles will increase fuel flow for both planes, but power remains at zero, thus it seems that fuel flow depends on thrust, not on power.
I'd think to maintain straight and level flight we need a fixed amount of lift, but the drag is just disturbing us. So we are looking for all combination of $C_L$ and speed so that the resulting lift is as much as we require to maintain straight and level flight. Then from those values we select the value that has minimum drag. This is the speed we need to fly, irrespective of the type of engine. Then we burn as much fuel as we need to generate the same amount of thrust as drag. This thinking of mine is obviously wrong, but why? What am I missing?
EDIT: While the answers I got seem to be perfectly logical to me and technically correct, it seems to me that they don't quite answer the question (or I am too stupid to make the connection). I guess I was not clear enough, so let me try to outline my mental picture so far:
In an Otto engine (for simplicity let's assume a single cylinder engine) a fuel-air mixture is injected into a cylinder, where it is compressed and then it is ignited. The fuel burns and thus the air becomes hot and wants to expand, but there is a piston in the way. Thus the pressure rises and exerts a force on the piston. This force (measured in Newton) causes the piston to move from TDC (top dead center) to BDC (bottom dead center). Thus a force is applied to over a distance (the distance between TDC and BDC), resulting in torque, measured in Newton-meters, equivalent to energy measured in Joule. In a perfect engine the amount of energy released that way would be the same as the amount of chemical energy contained in the fuel.
Increasing the amount of fuel that is injected into the cylinder will cause the the air in the cylinder to become hotter and expand more forcefully. The distance between TDC and BDC remain the same, but the force increases, resulting in increased torque.
But for power we need to know at which rate we are releasing the energy, so we divide the energy released during one stroke by the time it takes from the ignition to the next ignition. Thus we will get the power measured in Newton-meters per second or Joule per second or simply Watt.
From the above it quite clear that fuel consumption for a piston engine is directly connected to the power it produces.
What will happen if the propeller is feathered? In this case the power absorption of the propeller is zero and if it were not for friction the propeller would indefinitely continue to rotate with zero fuel flow. No force is needed to keep the rotation is progress. The piston would still move up and down, but no force is there and thus power is also zero.
Now let's unfeather the propeller. The rotating propeller will now move through the air like a wing. This creates an aerodynamic force pulling the plane forward, which we call thrust. The amount of thrust it develops depends on the angle of attack and the speed of the propeller blade (and air density, but we'll assume that this is constant). The speed of the propeller is obviously different at the root and the tip, just like the AoA, but we can work with an average AoA and an average speed.
Here my mental model starts to get confused. How is the energy transferred? I know must have something to do with out TAS, but I don't get a picture how that happens.
To simplify things I picture a plane on the ground. We use a rope to attach the tail of the plane to a force meter and attach the force meter to a wall. Now we can startup the engine and read the thrust on the force meter. I think we will all agree that if we increase fuel flow, the force meter will show an increase in thrust. But what happens if we now add some head- or tailwind, while keeping fuel flow constant? In my mental picture the force meter will still display the same thrust (as long as we adjust prop speed or prop pitch so that fuel flow remains constant). But that seems to be wrong. Why?

I might come back tomorrow with a more detailed answer, but very roughly with a propeller you're delivering constant power so thrust drops with speed as F=P/V, but with a turbojet you're delivering constant thrust so effective power increases with speed for the same throttle setting as P=FV.

"In case of propeller aircraft, the fuel flow rate is proportional to the power produced. Hence, the maximum endurance occurs at a point where the power is minimum." ... could this explain why Coast Guard C-130s would conduct searches with 2 engines shut down?

So as not to assume anything, do you really mean maximum endurance(time) or maximum range(distance)? Also it should be noted that for all the theory, a fair bit depends on specifics like selection of propeller pitch and weight load-balance. Jets are very inefficient at low power settings.

@MaxPower: I mean endurance (time)

Difference between engine types
That different engine types cause the optimum loiter speed to be different is due to the amount of air used for thrust creation. Propellers accelerate a lot of air by a little, while jets accelerate a little air by a lot.
Thrust is the difference between the exit impulse and the entry impulse of the air flowing through the engine rsp. propeller minus the installation drag (spill drag, cooling drag). If $v_0$ is the air speed ahead of the engine and $v_1$ that behind it, $\dot m_\text{air}$ the air mass flow and $\dot m_\text{fuel}$ the fuel flow, thrust $T$ is
$$T = (\dot m_\text{air} + \dot m_\text{fuel})\cdot v_1 - \dot m \cdot v_0 - D_\text{installation}$$
As long as $v_0$ is low, the speed difference $\Delta v = v_1 - v_0$ can be low and still thrust is high if enough air is involved. However, in order to fly fast, $\Delta v$ must be high enough so maximum thrust is achieved with a small air mass flow - using a bigger air mass would drive up engine mass and installation drag.
As a consequence, propellers produce declining thrust with increasing flight speed (and, consequently, increasing entry impulse) while in jets with their high exit speed thrust is approximately constant over speed in the subsonic region - it is high enough so that subtracting the entry impulse doesn't make much of a difference. Jet engines are also better in making good use of that entry impulse: By precompressing the air ahead of and in the intake, this ram effect raises the pressure level and mass flow inside the engine.
A piston engine consumes fuel in proportion to its speed (RPM) and air density, but independent of air speed. Thus, its power output is independent of flight speed. Thrust is power divided by speed, so this is another way to show why thrust must go down with increasing flight speed. Contrast this with jets, where precompression in the intake increases mass flow over flight speed, so fuel consumption will go up with speed. A typical high-bypass turbofan will consume twice the fuel flow per unit of thrust at cruise than it will at zero speed.
Maximum endurance speed
Why the optimum speed for maximum endurance differs with the thrust-over-speed behavior of the engine is explained in this answer. But what about gliders?
To stay airborne for the longest time requires to fly at the minimum sink speed. The "fuel" which keeps the glider flying is its potential energy $E_{pot}$ which is continuously converted into kinetic energy to compensate for what is used up by drag. So we can equate:
$$\frac{dE_\text{pot}}{dt} = m\cdot g\cdot \frac{dh}{dt} = m\cdot g\cdot v_z = D\cdot v$$
with $m$ = mass, $g$ = gravitational acceleration and $h$ = height. $v_z$ is the vertical speed which we aim to minimize.
If we call the flight path angle $\gamma$, we can write:
$$m\cdot g = -\frac{L}{\cos \gamma}\; \text{and}\; v_z = v\cdot \sin\gamma$$
For small values of $\gamma$, the approximations $cos \gamma = 1$ and $sin \gamma = \tan \gamma$ are valid. Together with the parabolic drag equation we can now write:
$$v_z = v\cdot\frac{c_{D0}}{c_L} + v\cdot\frac{c_L}{\pi\cdot AR\cdot\epsilon} = v^3\cdot\frac{c_{D0}\cdot\rho\cdot S_{ref}}{2\cdot m\cdot g} + \frac{1}{v}\cdot\frac{2\cdot m\cdot g}{\rho\cdot S_{ref}\cdot\pi\cdot AR\cdot\epsilon}$$
and set the derivative of $v_z$ over flight speed $v$ to zero:
$$\frac{dv_z}{dv} = 3\cdot v^2\cdot\frac{c_{D0}\cdot\rho\cdot S_{ref}}{2\cdot m\cdot g} - \frac{1}{v^2}\cdot\frac{2\cdot m\cdot g}{\rho\cdot S_{ref}\cdot\pi\cdot AR\cdot\epsilon}$$
$$= 3\cdot\frac{c_{D0}}{c_L} - \frac{c_L}{\pi\cdot AR\cdot\epsilon} \,\overset{!}{=}\, 0$$
This means that at the speed of minimum sink rate the induced drag must be three times as big as the zero-lift drag, which is different from the point of best glide speed, at which both are of the same size.
Nomenclature:
$c_{D0} \:$  zero-lift drag coefficient
$c_L \:\:\:$ lift coefficient
$S_{ref} \:$  reference area (wing area in most cases)
$v \:\:\:\:\:$  airspeed
$\rho \:\:\:\:\:$  air density
$\pi \:\:\:\:\:$  3.14159$\dots$
$AR \:\:$  aspect ratio of the wing
$\epsilon \:\:\:\:\:$  the wing's Oswald factor
$m \:\:\:\:$  the aircraft's mass
$g \:\:\:\:\:$  gravitational acceleration

EDIT: Your expanded question now asks what will happen to a stationary, piston-engine-powered aircraft in varying wind. Details now depend on the type of propeller which is fitted to the aircraft. Let's look at both possible cases and assume we change from no wind to a strong headwind:

Fixed-pitch propeller: With increasing headwind, the load on the propeller will go down, allowing the engine to run faster. If you keep the fuel flow constant, torque will go down, so the engine will settle at a higher speed where the load on the propeller blades is lower. Remember, power is also torque times angular velocity. Thrust will go down.
Constant speed propeller: Now the propeller will adjust pitch such that the engine speed is held constant. The blades will have a steeper angle with the propeller disc, and the resulting forces on them will point more sideways than before. This requires an overall reduction in the load on the propeller blades to keep the circumferential force component constant, and the force component perpendicular to the propeller disc will have to go down. Thrust will go down.

Either way, flying faster will result in less thrust. Please adjust your mental picture accordingly.

@ymb1 Hey - I upvoted your answer, don't delete it!

Thank you for your answer. While I believe that I understand what you are saying, I feel that the answer does not quite fit the question. I understand that propellers will produce less thrust at high TAS (on full throttle), but I think fuel flow will also reduce in this situation and thus this is no problem for endurance... I expanded my question...

@yankee: No, the fuel flow of piston engines is independent of speed. It only depends on throttle setting and air density. With jets this is different; their flow goes up with speed because precompression produces a higher mass flow.

I would enjoy very much a book (or a webpage) entitled The Collected Answers of Peter Kämpf...

@xxavier: Would you mind if that book is in German only?

@Peter Kämpf: Of course I wouldn't. On the contrary, I would very much appreciate a book/webpage entitled 'Die gesammelten Antworten von Peter Kämpf'...

The thing (most) the textbooks and these answers miss is that these idealized models of propulsors are  approximations -- and they aren't very good ones at that.
'Power producing engines' like piston engines (and most things that drive propellers -- are considered so because the maximum power is supposedly independent of velocity, and the fuel consumption is proportional to power -- $\dot{W}_f=\mathrm{BSFC} P$.  In order for that to be useful, BSFC needs to be relatively constant with velocity, altitude, and throttle.
'Thrust producing engines' like jet engines supposedly have maximum thrust independent of velocity and fuel consumption proportional to thrust $\dot{W}_f=\mathrm{TSFC} F_n$.  In order for this idealization to be useful, TSFC needs to be relatively constant with velocity, altitude, and throttle.
These assumptions underlay all of the best range, best endurance, and best climb -- occurring at best L/D, CL^3/2/CD, CL^1/2/CD, etc.  They also underlay the assumptions for the Breguet range equation for jets and props.
In reality, these approximations aren't very good.  IMHO, they are bad enough that textbooks should not be teaching them the way they do (without an appropriate critical eye).
In real engines, power and thrust available vary -- guess what, so does fuel / energy consumption.  This variation is complex.  This variation is very real and is important for aircraft performance.  The over-simplifed models are misleading.
Maximum endurance for jets and props is not different.  It is the speed and throttle setting at a particular altitude that produces the minimum fuel flow rate $\dot{W}_f$.  Period.  Full stop.
How you calculate $\dot{W}_f$, how it varies with speed and throttle, and how close that can be approximated by simplified models varies.

Gas turbine (jets) as well as turbofans, are more efficient when the aircraft is moving at speed because the ram effect of air moving through the engine is increased (much like a supercharger or turbocharger on a piston engine) There is no equivalent effect on a propeller driven aircraft. Because of this effect a jet engined plane loves to fly at high altitude 30000-40000 feet at high speed. All passenger planes do this. If they are forced down to say 20,000 ft fuel consumption goes up because of drag and it’s not economical.


text:
What do winglets do to increase aircraft performance?
It seems like a lot of the newer airliners have winglets or wing fences. How do they improve aircraft performance?

they reduce the wingtip vortex and the associated drag by deflecting the air that wants to escape over the tip back down
(source: wikipedia)

For lots more detail than you ever wanted to know about wingtip vorticies, their induced drag, and wingtip devices to counteract them check out Sighard Hoerner's book Fluid Dynamic Drag. The ZenithAir website also has some more accessible explanations (about 1/3 of the way down under "Wing Tips").

Wingtip vortices do not create drag; they are a consequence of the process that creates both lift and drag - see here and here. See page 4.6 in the latter for a discussion of this myth, and page 4.8 for an explanation of induced drag reduction as a consequence of how a non-planar wing has a different (flatter) least-induced-drag lift distribution than a planar wing of the same span.

Wing generates lift by creating area of higher pressure below and area of lower pressure above.
At the wing tip, some air flows around the tip, reducing the pressure differential and thus the efficiency of the wing. This is called transverse flow. The wing tip reduces this flow, thus improving the efficiency.
The effect of wing tip is actually similar to making the wing a bit longer, but since the wing tip does not produce lift itself, it is less stressed and therefore can be lighter, even though the extension would be about 1/3 shorter for comparable efficiency.

Winglets, raked wingtips, fences, sharklets all do the same thing; reduce drag.  When the wing is working hard (high AoA), it will produce a lot of this kind of drag since vortices of all kinds will be stronger.  Winglets makes the passage of air from the bottom of the wing to the lower pressure top of the wing more difficult, thus reducing induced drag.
The big manufacturers explored offering winglet retrofit kits to their customers but at the time, it wasn't economical since fuel was so cheap.  These days, however, the retrofits pay for themselves fairly quickly even on airplanes wings that have a fairly high aspect ratio.  Every 737/757/767 guy that I talk to says they get in the neighborhood of a 4-6% boost in fuel savings.
On an tangentially related topic, the hump from the satellite internet antennas has a negligible impact on performance, less than 1% according to the few guys that I talked to.

Lol, what? 1% of fuel for just an antenna? No thanks!

OK, 5 years late here. @Alexus, agreed, 1% is a lot but he said less than 1% and the money they can make in exchange for the in-flight web connection is apparently enough to offset. Probably more likely that it is partially baked into ticket prices which is why it is not available on budget airlines.

Basically what happens with an airplane without a wingtip is the high pressure area comes over the lower pressure area and creates a giant vortex called a wing tip vortex and the winglet reduces the strength of the vortex reduces drag, increasing lift, and increasing the aircrafts range.

an airplane without a wingtip has infinite wings

or circular wings


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How do Airbus planes flare in normal operation?
In reading the reports and Sully's comments regarding the flight that was ditched in the famous events of US Airways Flight 1549, I am wondering how the Airbus control system works to flare an airplane under normal flight conditions.
Sully stated that since the automated system did not let him flare as much as he wanted to, he was unable to reduce vertical speed further.
Is it that under normal flight, the aircraft has forward speed to generate lift, and the descent rate is thus less? I thought (I'm not a pilot) that it was desirable to basically stall by flaring at the last moment in normal powered landing, thus sticking it to the runway for traction, wheel spin-up, etc.

He would not have wanted to flare into a stall. At stall the lift drops dramatically and would increase vertical speed. It wouldn’t let him flare all the way to the point before the stall.

He was "mushing" towards the water and was interested in hitting the water aft first but the flight computer would not allow him raise the nose just before impact.

@TomMcW, not necessarily. Stall is not always abrupt, and in some circumstances one may want to use up all the margin up to the actual stall. Even if you go slightly beyond stall (top of the lift curve), but you are just a few feet off the water, vertical speed will not have time to develop significantly, yet every moment you are winning time, i.e. reducing forward speed.

The only good thing about this question is the OP's handle. Blondie reference for the win! :)   (If you stall in the flare, you are doing it wrong if you are flying an Airbus ... an indeed in most other fixed wing aircraft). 8^o desirable to basically stall by flaring at the last moment ... no, that isn't the desire. You want to be flying as you touch down, and stalling isn't flying - it's falling.  ;)

tough crowd over here at Aviation. Guess I'll go watch Top Gun and cry...

You don't do full stall landings in airliners.  You arrive over the threshold at reference speed (the full flaps approach speed), ease the thrust off (how soon depends on the airplane, but you need to be at idle before touchdown) and start to flare but only enough to reduce the sink rate to close to zero and then you tease the pitch to maintain a gentle sink until you touch.   If you wanted to do a full stall type landing, there is sufficient energy still available at reference speed that you would have to hold it off and float halfway down the runway, and you'd touch down with the stick shaker going.  Not a desirable outcome in a 150,000 lb soda can.
If you arrived over the threshold too much below reference speed, there is no energy reserve to save yourself if you are sinking too fast and there goes your tires, if you're lucky.
Water landings often don't go so well if the tail is too low (if the tail touches in too nose high an attitude, it tends to bounce upward and drive the nose over and things can go south).  So it's possible the FBW system actually protected them by limiting Sully's ability to pitch up.  They seemed to have touched down at an ideal attitude, getting a nice ski effect from the nacelles without the tail hitting too early.
In the famous Airbus airshow incident, the airplane descended gently into the trees as it struggled along way on the back side of the power curve (in that regime, the only option is to descend).  Had the crew more pitch up authority available to them, it is possible they would have semi-stalled into the trees at a much higher sink rate. That was Airbus's response to the criticism over the FBW's pitch limiting behaviour.

Strictly speaking, descending is not the only option on the back of the power curve. You just increase thrust, if available. Most agile aircraft will push through stall with ease. So, hypothetically, if you need to win time to spin up the turbines, it may make sense to go beyond the protected angle. Or, maybe it's better to go down with higher vertical speed but lower forward speed: still, less energy. It'd rather leave this decision to the pilot.

The Airbus's pitch attitude was 15 degrees at the time they applied TOGA, so the wing with takeoff flap would have been on the edge of a stall.  There was no pitch margin left and the only thing keeping them in the air was ground effect.  Key bit: if it was a conventional airliner with a stick shaker and pusher, what would have happened is the stick shaker would've been going as they mushed along, then as they kept pulling, the pusher would have fired and they would've nosed into the ground instead of settling in a steady state.  So yes I think the computers saved their asses.

(but you need to be at idle before touchdown) Not in the MD-80 haha. At heavy weights with lots of airplane nose up "ANU" trim with the direct cable elevators if you pulled power to idle over the runway and she started to sink you would pull back on the elevator only for nothing to happen! Most left the power in and then swiped it at touchdown to avoid this. Only jet I've ever known where landing "power on" was encouraged.

Below 50ft radio altitude Airbus flight controls go to Flare mode. The key change there is that vertical control (stick front/back) changes from normal pitch rate close to conventional elevator movement.
This means that while in normal flight, i.e. cruise, holding the stick back at a certain position you would see constant nose-up movement on a given rate.  On Flare mode you are able to control (and maintain) pitch attitude more precisely, as the stick input will result in elevator deflection rather than pitch rate.
Flare mode isn't available after loss of radio altimeters.
Also after a dual engine failure, the aircraft would have most likely been in Direct law, meaning that all automatic trims are lost.  This might impact elevator efficiency, since the horizontal stabilizer would not automatically move.


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Why does drag coefficient decrease with Mach number in the supersonic regime?
I'm trying to understand why the drag coefficient decreases in the supersonic regime with Mach number. While it is easy to show this using supersonic, potential flow theory, I'm looking for a more physical explanation.
My initial thought is, as freestream Mach number increases, the shocks should become stronger (even though they are more oblique, the normal component to the shock is still larger). Where is this train of thought going wrong?

Read this and then let me know if anything is still unclear.

The answer indicates why wave drag occurs, but I'm still a little unclear on why it would decrease with Mach number. It doesn't really dive into why the drag would decrease, just that drag would occur as Mach number increases to sonic velocity and past it.

With speed well above Mach 1, the change in density is large enough to let the aircraft pass. Only around Mach 1 is the contraction of the streamlines low because the change in pressure and density compensate each other. In fully supersonic flow the change in density is dominant and makes it easier again for the aircraft to squeeze through.

As an airfoil goes from subsonic to transonic to supersonic flow, shocks form on the surface of the wing and then move off of the wing.
Look at the images here.
Most of the drag during the transonic drag rise is actually due to shock induced boundary layer separation.  This drag is not technically wave drag.
At higher Mach, there is now a bow shock off the front of a blunt leading edge -- or an oblique shock off of sharp leading edges.  There will also be a shock at the trailing edge.  The flow separation on the airfoil surface is gone and the drag is reduced.

From this answer we learn that an increase in flow speed is always coupled to a decrease in density. The ratio between both changes grows with the square of flow speed and equals unity when Mach equals the inverse square root of the ratio of specific heats.
Why is this important? For an airplane to fly through air, this air has to make way for the airplane. At subsonic speed this happens by an increase of local flow speed while at high supersonic speed the air slows down, getting denser. In both cases the air will temporarily occupy less volume so there is space for the airplane to pass.
Think of the air flowing past the airplane as flow through tubes. In order to make way for the airplane, those tubes must shrink in diameter around the airplane so it can fit in. They will do so at subsonic and high supersonic speeds, but much less around Mach 1. That is the reason for the maximum in the drag coefficient between Mach 1 and 1.2.


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What is the airplane in this video?
What is the airplane in this video?
(Video is intentionally set to start at 1:27; that is shortly before the first appearance of the airplane in the video.  After 1:50 there are no clear views of the exterior of the airplane, and only extremely limited views of the interior, so there is little need to watch beyond that point, unless you are into the music.  At 2:12-2:15 there is a brief low-resolution shot that includes a good portion of the control panel; nothing of significance after that.)

It's a Pacific Aerospace P-750 XSTOL


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Is this photo of an aircraft in a flood genuine?
I stumbled upon this picture somewhere on the internet:

I don't believe an aircraft would be able to take off like that; I also don't expect a normal landing be allowed if this were on a runway. I suppose, in case of emergency, it could possibly land into such water. Generally, I think that either this picture is fake or the aircraft was parked at a stand before the area was flooded.
Unfortunately, as the picture was so small, it's virtually impossible to see any details, and thus difficult to analyse for "fakeness". Could this picture possibly be real?

You might be able to go through it if the engine is already running...

One clue that it is a fake is the reflection of the taxiway center-line under the nose of the aircraft.

How can "is this real" be an unclear question?

Yes, the photo is a "Fake".
The photo was taken by Gary Watt of Aberdeen Scotland and then photo-shopped using the  "Flood Filter".
PPRUNE explanation

Hi all
  Amazing how this image gets around after posting it to someone some months back.
  Yes, it's a fake. It's my photograph taken at ABZ and the 'Flood Filter' done in Photoshop.
  Great tool and can produce some fantastic effects.
  For all the doubters, the engines ARE running as it was taxiing to 34 for departure.
  I can still confirm it\'s fake and if anyone else thinks different, ye need yer heids examined
  Cheers
  Gary Watt

I found a larger version of the photo on this web page.  I also found some comments about it here.  Quoting from what someone said on that site: "There is no standing water anywhere else in the picture, there appear to be pilots in the cockpit, the leading edge flaps are set and there appears to be a lot of heat haze behind the starboard engine."
It's a fake.


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Is Touring Motor Glider included in the term "aeroplanes"?
The EASA definition goes as follows:
‘Aeroplane’ means an engine-driven fixed-wing aircraft heavier than air, that is supported in flight by the
dynamic reaction of the air against its wings.
‘Aircraft’ means a machine that can derive support in the atmosphere from the reactions of the air other
than the reactions of the air against the earth’s surface.
To me a TMG could go under that definition but not sure though...
Basically I'd like to know whether the 15 hours that are needed to go from LAPL(a) -> PPL(a) can be done on TMG
https://www.caa.co.uk/General-aviation/Pilot-licences/EASA-requirements/PPL-SPL-BPL/PPL-(A)-requirements/

Wow that's a weird definition of "aircraft".  I guess they are trying to say "out of ground effect".  But I think you could make a case that the earth's surface IS playing some role in supporting an aircraft, helicopter, or balloon, even if flying at high altitude.

@quiteFlyer the definition of aircraft is such that wing-in-ground effect craft and hovercraft are excluded, but balloons, aeroplanes, gliders, helicopters etc are included. The earth's surface may be playing some role at low level flight, but it does not have to.

From an Easa Certification point of view, motorgliders, motor assisted (turbo) gliders and regular gliders are all certificated(nowadays) under CS 22, while light aircraft are certification under 23.. CS  is Certification Standard , equivalent to FARs .

To my understanding of Part-FCL, yes.
While aeroplanes is a broad term used in the FCL, they use it in some cases to include both, SEP and TMG, e.g.

FCL.105.A LAPL(A) —Privileges and conditions (a) The privileges of the holder of an LAPL for aeroplanes are to act as PIC on single-engine piston aeroplanes-land or TMG with a maximum certificated take-off mass of 2 000 kg or less, carrying a maximum of 3 passengers, such that there are never more than 4 persons on board of the aircraft.

and hence

FCL.210.A (b)  Specific requirements for applicants holding an LAPL(A). Applicants for a PPL(A) holding an LAPL(A) shall have completed at least 15 hours of flight time on aeroplanes after the issue of the LAPL(A), of which at least 10 shall be flight instruction completed in a training course at an ATO. This training course shall include at least 4 hours of supervised solo flight time, including at least 2 hours of solo cross-country flight time with at least 1 cross-country flight of at least 270 km (150 NM), during which full stop landings at 2 aerodromes different from the aerodrome of departure shall be made.

Source: https://www.easa.europa.eu/sites/default/files/dfu/Part-FCL.pdf


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what does upwind engine and down wind engine mean?
can anyone illustrate with a picture what does upwind engine and downwind engine mean?
for example, i came across this - “The outer engines are critical. If there is any crosswind then the upwind outboard engine is the most critical. With a twin jet the upwind engine is critical for the same reason in a crosswind.”

Up wind is the same as windward, or into-wind side. If the crosswind is from the left, then the left engine is critical, because if it fails, you have the crosswind from the left, AND the asymmetric thrust from the right engine(s) trying to drive the nose to the left during the takeoff phase while the wheels are in contact with the surface.
Once you are clear of the ground, you can align the plane into the airflow with rudder, and if you do nothing else you will start to drift downwind as the crosswind carries you laterally, or, if you are able to think about it, you can do a slight turn into wind to establish a crab angle to maintain the runway axis on the climbout.  But in any case, a few seconds after you are airborne, the crosswind is no longer relevant from a critical engine aircraft control standpoint.


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What does this black lever next to the overhead panel on the Airbus 320 do?
What does this do?
It's located next to the overhead panel of an A320 cockpit.

@Bianfable it appears to be to the bottom-right hand side of the overhead panel. Next to the ventilation controls.

I’d guess it is the ventilation control. Probably opens and closes the louvers for that overhead vent.  Interesting that the pilot doesn’t seem to have one too.

Prob due to the HUD

Why not give it a yank and see what happens?

It's a secret and Airbus does not want you to know.)
Here is the cockpit general arrangement view from the Airbus Flightdeck and Systems briefing for Pilots which labels the visor and rope storage compartment, but the pdf in every version that I looked at only showed descriptions for controls but the graphic did not render. Anywhere.

Next I looked at a flight crew operating manual, where the control is not even pictured but everything around it is. This may be only useful in determining what it is not.

I looked at photos including this one from a German Wings A320 where there are two, so it cannot be that critical a control if this has two, your photo has one, some apparently have none, and nowhere is it documented.

The best answer that I found comes from the cabin crew operators manual which shows ventilation ducts going right through that area on both sides of the overhead panel, terminating in a plenum above each of the pilot and copilot windscreens and showing airflow from those plenums across the windscreens. Since the control is a variable +/- from your picture, it stands to reason that it is a simple airflow control for defogging the windscreen.

+1 for opening sentence --


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What is this sound on a B737-800 Takeoff?
Does anyone know what this sound is at 12:56 on a B737-800 takeoff?
The aircraft is N822SY, the airport is Denver Intl. Airport, and the runway is 34L.
Thanks and have a great day!

A time stamp and description would be helpful.

timestamp: 12:56

It's the main gear wheels vibrating as they spin down.  As the RPM of the wheel assembly drops, the wheel will pass through resonant frequencies and if there are small amounts of imbalance, the wheel will vibrate while passing through a specific speed range that is resonant with gear leg and structure, transmitting the vibrations into the fuselage and rattling interior panels.
At some point the brake system is applied automatically to stop the main wheels from spinning, but prior to that you will get those little vibrations. The nose gear also can vibrate while spinning down, and will do so until a friction brake in the gear well rubs the tires to a stop following retraction.
Airline tires are only statically balanced, a bit like motorcycle tires are, and it's not unusual for them to vibrate when suspended clear of the ground spinning at high speed, whereas when on the ground with weight on the gear, the imbalance is not noticeable because it's heavily damped by the gear and structure pushing down hard on the wheels.

Given that some bridges were subject to catastrophic resonance failure and modern bridge designs have shifted resonance frequencies, so it can not be excited by wind, is there a reason why it's allowed in air craft wheels? Seems like it could cause a lot of stress. Although I don't know how it compares to stress when touching down.

@infinitezero: someone might come with a more technical answer, but from a practical point of view, millions and millions of flights without incident seem like a good proof that this is not an issue.

The level of balancing required to prevent any spindown shaking on 5 or 6 foot tire assemblies would require dynamic balancing machines that few airlines would be willing to pay for and I've never seen an airliner-size dynamic balancer machine like is used for cars.  Everybody uses static balancers that hold the assembly with its axle horizontal, like motorcycle tire balancers.  Spindown shaking happens on just about all airplanes when you take off, some more noticeable than others.  It can be especially bad on Cessnas with the spring steel gear, since the gear acts like a harmonic amplifier.

@infinitezero it's pretty hopeless to eliminate all resonances, short of filling all cavities with a heavy viscous liquid or so. The important thing to get rid of are those resonances that can actually excited without bounds. The resonances in question here are apparently not excited by wind, but only by the wheels – and those can only provide so much energy, because the more resonance there is the quicker the wheel also slows down – so you can always only get a small fraction of the wheel's rotational-kinetic energy into the resonance mode. Much less stress than during a landing.


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How do canard flaps work?
Before examining this XB-70 picture
Wikipedia XB-70 photo

I thought that all canard were one single flight control surface (as seen on many fighters such as the Eurofighter Typhoon), and that there functionning were comparable to stabilators.
But the XB-70 exhibit something that looks like large flaps at the trailing edge of its canard. Apparently, it is not the only airplane with this feature (e.g. the Saab 37 Viggen is also equipped with similar canard).
What is this canard flap? How does it work? Is there any pro/cons of this configuration compared to a all moving canard (I don't know if this can be called stabilator as this is not located on the tail)?

It still could be a stabilizer. Based on the picture it looks to be that the canard is not used to generate vortex lift like the purpose of the Eurofighter Typhoon canard. It could act in the same way as a stabilizer and those are elevators. Since its a supersonic aircraft it has no stabilizer because there is too much drag so the canard is just a short stabilizer. Good question.

The canard flap is like any other control surface in the trailing edge of a canard. The canard flaps are there mainly for trimming the aircraft.
At transonic speeds, the center of pressure of the aircraft moves rearward, resulting in instability. This could be countered by trimming with the elevons; however, this increases the angle of attack (the elevon deflection in XB70 was between $0^{\circ}$ and $10^{\circ}$)- resulting in increased drag. The small flaps in the XB70's canard was used to counter the movement of center of pressure without changing the main wings AoA, which eliminated the speed and range penalty due to increased drag.
Another advantage is during the landing. The landing attitude of delta aircraft is usually steep and fast compared to other aircraft. The deflection of canard flaps in the XB-70 pitched the nose up, enabling the elevons to be operated as trailing edge flaps- this helped the to keep the landing attitude shallower and speed, lower.
Actually, in case of XB-70, the canard was exclusively a trimming device- the longitudinal control was provided solely by the elevons.
During landing, the use of flaps cause a negative pitching moment- In aircraft with horizontal stabilizers, flap deployment may increase the downward airflow on the horizontal tail, producing a pitch up moment. in case of aircraft without stabilizers, it is not possible. In order to trim the aircraft, the canard flaps in Saab Viggen were deployed; in case of Beechcraft starship, a variable sweep canard was used for the same purpose.
The main disadvantage of having flaps in canard was the increased complexity. The canard can in itself be used to provide the required trim; the flaps are used only when the canard is highly loaded and may not have sufficient spare lifting capacity to accommodate large movements of the centre of pressure or of the centre of gravity. Note that in case of XB-70, initially, the canard was fixed and the flaps were adjustable; later the canards were made trimmable with the flap deflection set at two positions- $0^{\circ}$ and $20^{\circ}$.
References: WarbirdTech: North American XB-70A Valkyrie by Dennis R. Jenkins and Tony Landis; Summary of Preliminary Data Derived from the XB-70 Airplanes by William H. Andrews, NASA Technical Memorandum TM X-1240


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What is this black streak on the F-35B?
I was seeing this image and noticed a black streak behind a vent on the fuslage of the F35. What is this vent used for?

Details between the auxiliary intake and the lift fan (source) - Model - Top view

I edited your answer for clarity. Your use of the word “smoke” was not really correct. None of the photos online show smoke coming out of the vent, but some do show a black residue on the fuselage. It appears that that some of the engine exhaust is vented from that hole, and it leaves a sooty stain on the aircraft. It would be nice to see what the answer is.

@MikeSowsun Ok, Thank you Mike.